Physics A2 signals

[yasser37]

Hi all

I have a problem with these questions
I don't know how to do them
the ones that come in section B about signal to loss ratio and all. the 10 log(p-out/P-in)

if there's a video or a page that explains these questions I would be thankful
my exam after about 36 hours so I need something clear to explain it

Thanks

[yasser37]

please anyone 

[Sue T]

ill explain but with my own words - other than the application booklet ive got nowhere else 2 tell u 2 refer to - so shall i go ahead and xplain my way?

[yasser37]

I'd be thankful mate 

[yasser37]

bump 

[Sue T]

ok so you have a general rule: decibels = 10lg(P₁/P₂)
which can be used 4 any quantity involving power ratios such as gain, attenuation, or signal-to-noise ratios.

lets take up gain first. its basically when a smaller power is input and a larger power is output. Now for as far as you no, gain has always been output/input ryt? (check the eqns involving amplifiers)
so   gain               is         ratio of output/input
so   gain in decibels = 10lg(P_output/P_input)

now if you have attenuation. It describes a larger input going in and a smaller output resulting due to loss of power along the way. so it describes that ratio of ur input to output (this is proven by the fact that wenever they give u an attenuation it is a whole number - meaning ur numerator is more than your denominator - so large over small = attenuation)
so attenuation               is    ratio of input/output
so attenuation in decibels = 10lg(P_input/P_output)
again remember that its large/small

now signal-to-noise ratio speaks for itself ryt?
so signal-to-noise ratio                 is      ratio of signal to noise :p
so signal-to-noise ration in decibels = 10lg(P_{output or signal}/P_noise)

sometimes attenuation comes as attenuation per unit length - so u divide my length - simple!
attenuation in decibels = 1/L 10lg(P_input/P_output)
 ill find an example to do

[Sue T]

look at mj10_42 Q12
attenuation per unit length = 1.6 dBkm^-1
minimum signal-to-noise ratio = 25dB
noise power = 6.1 x 10^-19 W
input power to cable = 6.5 x 10^-3 W

min sig-to-noise ratio = 10lg(P_min-output/P_noise)
25 = 10lg (P_min-out/6.1 x 10^-19)
so P_min-out = 1.93 x 10^-16

so if the minimum power output possible was just found, you find the maximum possible length of cable (where any length of cable larger than that would result in a smaller output power that is not wanted)
attenuation per unit length = 1/L_max 10lg(P_input/P_min-output)
1.6 = 1/L_max 10lg(6.5x10^-3/1.93 x 10^-16)
eventually L_max = 84 km
so if the length of the wire between the two towns was 75km - no need 4 amplifiers

[yasser37]

Dude.I love you
u made it seem simple
+rep